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(B) The electric flux $\phi$ through a surface is given by the projection of the area vector onto the electric field direction,or $\phi = \vec{E} \cdo

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$EhR$

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(B) The electric flux $\phi$ through a surface is given by the projection of the area vector onto the electric field direction,or $\phi = \vec{E} \cdot \vec{A}$.For a cone placed in a uniform electric field $\vec{E}$ parallel to its base,the flux entering the cone is equal to the flux passing through the projection of the cone onto a plane perpendicular to the electric field.The projection of the cone onto a plane perpendicular to the electric field is a triangle with base $2R$ and height $h$.The area of this triangular projection is $A = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes (2R) 	imes h = Rh$.Since the electric field lines enter the cone through this projected area,the electric flux entering the cone is $\phi = E 	imes A = E 	imes (Rh) = EhR$.

$A$ cone of base radius $R$ and height $h$ is located in a uniform electric field $\vec{E}$ parallel to its base. The electric flux entering the cone is

Similar Questions

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